ANNALES DE LA FACULTÉ DES SCIENCES
Mathématiques
STEPAN YU.OREVKOV Automorphism group of the commutator subgroup of the braid group
Tome XXVI, no 5 (2017), p. 1137-1161.
© Université Paul Sabatier, Toulouse, 2017, tous droits réservés. L’accès aux articles de la revue « Annales de la faculté des sci- ences de Toulouse Mathématiques » (http://afst.cedram.org/), implique l’accord avec les conditions générales d’utilisation (http://afst.cedram. org/legal/). Toute reproduction en tout ou partie de cet article sous quelque forme que ce soit pour tout usage autre que l’utilisation à fin strictement personnelle du copiste est constitutive d’une infraction pénale. Toute copie ou impression de ce fichier doit contenir la présente mention de copyright.
cedram Article mis en ligne dans le cadre du Centre de diffusion des revues académiques de mathématiques http://www.cedram.org/ Annales de la faculté des sciences de Toulouse Volume XXVI, no 5, 2017 pp. 1137-1161
Automorphism group of the commutator subgroup of the braid group (∗) Stepan Yu. Orevkov (1)
0 ABSTRACT.— Let Bn be the commutator subgroup of the braid 0 group Bn. We prove that Aut(Bn) = Aut(Bn) for n > 4. This answers a question asked by Vladimir Lin.
0 RÉSUMÉ.— Soit Bn le groupe dérivé du groupe de tresses Bn. On 0 montre que Aut(Bn) = Aut(Bn) pour n > 4, ce qui répond à une question posée par Vladimir Lin.
1. Introduction
Let Bn be the braid group with n strings. It is generated by σ1, . . . , σn−1 (called standard or Artin generators) subject to the relations
σiσj = σjσi for |i − j| > 1 ; σiσjσi = σjσiσj for |i − j| = 1. 0 Let Bn be the commutator subgroup of Bn. Vladimir Lin [16] posed a prob- 0 lem to compute the group of automorphisms of Bn. In this paper we solve this problem.
Theorem 1.1. — If n > 4, then the restriction mapping Aut(Bn) → 0 Aut(Bn) is an isomorphism. ∼ Dyer and Grossman [5] proved that Out(Bn) = Z2 for any n. The only nontrivial element of Out(Bn) corresponds to the automorphism Λ −1 defined by σi 7→ σi for all i = 1, . . . , n − 1. The center of Bn is gener- 2 Qn−1 Qn−i ated by ∆ where ∆ = ∆n = i=1 j=1 σj is Garside’s half-twist. Thus
(*) Reçu le 11 novembre 2015, accepté le 5 juillet 2016. (1) Institut de Mathématiques de Toulouse, Université Paul Sabatier, 118 route de Narbonne, 31062 Toulouse, France — Steklov Mathematical Institute, 8 Gubkina St., 119991 Moscow, Russia — [email protected] I am grateful to the referee for many very useful remarks. Article proposé par Jean-Pierre Otal.
– 1137 – Stepan Yu. Orevkov
∼ 2 Aut(Bn) = (Bn/h∆ i) o Z2. For an element g of a group, we denote the inner automorphism x 7→ gxg−1 by g˜.
0 Corollary 1.2. — If n > 4, then Out(Bn) is isomorphic to the dihedral group Dn(n−1) = Zn(n−1) o Z2. It is generated by Λ and σ˜1 subject to the 2 n(n−1) defining relations Λ =σ ˜1 = Λ˜σ1Λ˜σ1 = id.
0 For n = 3, the situation is different. It is proven in [11] that B3 is a free −1 −1 group of rank two generated by u = σ2σ1 and t = σ1 σ2 (in fact, the free 0 −1 base of B3 considered in [11] is u, v with v = t u). So, its automorphism group is well-known (see [18, §3.5, Theorem N4]). In particular (see [18, Corollary N4]), there is an exact sequence
0 ι 0 π 1 / B3 / Aut(B3) / GL(2, Z) / 1
0 where ι(x) =x ˜ and π takes each automorphism of B3 to the induced auto- 0 2 morphism of the abelianization of B3 (which we identify with Z by choos- −1 −1 ing the images of u and t as a base). We have σ˜1(u) = t u, σ˜2(u) = ut , −1 −1 σ˜1(t) =σ ˜2(t) = u, Λ(u) = t , Λ(t) = u whence 1 1 0 −1 π(˜σ ) = π(˜σ ) = , π(Λ) = . 1 2 −1 0 −1 0
0 ∼ Thus, again (as in the case n > 4) the image of Aut(B3) in Out(B3) = 0 GL(2, Z) is isomorphic to D6 but this time it is not the whole group Out(B3).
Let Sn be the symmetric group and An its alternating subgroup. Let µ = µn : Bn → Sn be the homomorphism which takes σi to the transposition 0 0 (i, i + 1) and let µ be the restriction of µ to Bn. Then Pn = ker µn is the 0 0 0 group of pure braids. Let Jn = Bn ∩ Pn = ker µ . Note that the image of µ is An. The following diagram commutes where the rows are exact sequences and all the unlabeled arrows (except “→ 1”) are inclusions:
0 0 µ 1 / Jn / Bn / An / 1 (1.1)
µ 1 / Pn / Bn / Sn / 1
Recall that a subgroup of a group G is called characteristic if it is invariant under each automorphism of G. Lin proved in [15, Theorem D] that Jn is a 0 characteristic subgroup of Bn for n > 5 (note that this fact is used in our proof of Theorem 1.1 for n > 5). By Theorem 1.1, this result extends to the case n = 4.
– 1138 – Automorphism group of the commutator subgroup of the braid group
0 Corollary 1.3. — J4 is a characteristic subgroup of B4.
0 Note that J3 is not a characteristic subgroup of B3. Indeed, let ϕ ∈ 0 −1 Aut(B3) be defined by u 7→ u, t 7→ ut. Then ut ∈ J3 whereas ϕ (ut) = t 6∈ J3.
2. Preliminaries
Let e : Bn → Z be the homomorphism defined by e(σi) = 1 for all 0 i = 1, . . . , n. Then we have Bn = ker e.
2.1. Groups
For a group G, we denote its unit element by 1, the center by Z(G), the commutator subgroup by G0, the second commutator subgroup (G0)0 by G00, and the abelianization G/G0 by Gab. We denote x−1yx by yx (thus x˜(yx) = y) and we denote the commutator xyx−1y−1 by [x, y]. For g ∈ G, we denote the centralizer of g in G by Z(g, G). If H is a subgroup of G, then, evidently, Z(g, H) = Z(g, G) ∩ H. Lemma 2.1. — Let G be a group generated by a set A. Assume that there exists a homomorphism e : G → Z such that e(A) = {1}. Let e¯ be the induced ab homomorphism G → Z. Let Γ be the graph such that the set of vertices is A and two vertices a and b are connected by an edge when [a, b] = 1. If the graph Γ is connected, then (ker e)ab =∼ kere ¯. Proof. — Let K = ker e. Let us show that G0 ⊂ K0. Since K0 is normal in G, and G0 is the normal closure of the subgroup generated by [a, b], a, b ∈ A, it is enough to show that [a, b] ∈ K0 for any a, b ∈ A. We define a relation ∼ on A by setting a ∼ b if [a, b] ∈ K0. Since Γ is connected, it remains to note that this relation is transitive. Indeed, if a ∼ b ∼ c, then [a, c] = [a, b][bab−2, bcb−2][b, c] ∈ K0. Thus G0 ⊂ K0 whence G0 = K0 and we obtain Kab = K/K0 = K/G0 = kere ¯. 00 0 Remark 2.2. — The fact that Bn = Bn for n > 5 proven by Gorin and Lin [11] (see also [15, Remark 1.10]) is an immediate corollary of Lemma 2.1. n−1 Indeed, if we set G = Bn and A = {σi}i=1 , then Γ is connected whence 0 00 ab ∼ 00 0 Bn/Bn = (ker e) = kere ¯ = {1}. In the same way we obtain G = G when G is an Artin group of type Dn (n > 5), E6, E7, E8, F4, or H4.
– 1139 – Stepan Yu. Orevkov
2.2. Pure braids
2 Recall that Pn is generated by the braids σij, 1 6 i < j 6 n, where −1 −1 σij = σji = σj−1 . . . σi+1σiσi+1 . . . σj−1. For a pure braid X, let us denote the linking number of the i-th and j-th strings by lkij(X). If X is presented by a diagram with under- and over-crossings, then lkij(X) is the half-sum of the signs of those crossings where the i-th and j-th strings cross. Let Aij 2 ab be the image of σij in Pn . We have, evidently, γ lkγ(i),γ(j)(X) = lki,j(X ), for any X ∈ Pn, γ ∈ Bn (2.1)
(here γ(i) = µ(γ)(i) which is coherent with the interpretation of Bn as a mapping class group; see Section 4.1). ab It is well known that Pn is freely generated by {Aij}16i
3 3 2 2 2 (2) E3 = {u¯t,¯ t¯u,¯ u¯ , t¯ } and E4 = E3 ∪ {c¯ , w¯ , (¯cw¯) } are free bases of ab ab J3 and J4 respectively; u, t, w, c are defined in the beginning of Section6. ab (Here and below x¯ stands for the image of x under the quotient map Jn→Jn .) ab ab (3) Let pn : Jn → Pn , n = 3, 4, be induced by the composition Jn → ab Pn → Pn . Then n X X o 3 3 im pn = xijAij | xij = 0 , ker pn = hu¯ , t¯ i.
– 1140 – Automorphism group of the commutator subgroup of the braid group
Proof. — (1). The result is obvious for n = 2 and it follows from Lem- ma 2.3 for n > 5. For n = 3, the result follows from the following argument proposed by 0 ∼ 1 1 the referee. We have B3 = π1(Γ) where Γ is the bouquet S ∨ S . Since 0 ab ∼ ˜ ˜ |B3/J3| = 3 (see (1.1)), we have J3 = H1(Γ) where Γ → Γ is a connected 3-fold covering. Then the Euler characteristic of Γ˜ is χ(Γ)˜ = 3χ(Γ) = −3 ˜ whence rk H1(Γ) = 4. ab The group J4 can be easily computed by the Reidemeister–Schreier 0 0 method either as ker µ using Gorin and Lin’s [11] presentation for B4, or as ker(e|P4 ) using Artin’s presentation [1] of P4. Here is the GAP code for the first method: f:=FreeGroup(4); u:=f.1; v:=f.2; w:=f.3; c:=f.4; g:=f/[u*c/u/w, u*w/u/w*c/w/w, v*c/v/w*c, v*w/v/w*c*c/w*c/w*c/w*c]; u:=g.1; v:=g.2; w:=g.3; c:=g.4; # group B’(4) according to [11] s:=SymmetricGroup(4); t1:=(1,2); t2:=(2,3); t3:=(3,4); U:=t2*t1; V:=t1*t2; W:=t2*t3*t1*t2; C:=t3*t1; # U=mu(u),V=mu(v),... mu:=GroupHomomorphismByImages(g,s,[u,v,w,c],[U,V,W,C]); AbelianInvariants(Kernel(mu)); # should be [0,0,0,0,0,0,0]
t tt u t u u u
Figure 2.1. The graphs Γ and Γ˜ in the proof of Proposition 2.4
(2) for n = 3. In Figure 2.1 we show the graphs Γ and Γ˜ discussed above. ˜ We see that the loops in Γ represented by the elements of E3 form a base of ˜ H1(Γ).
(3) for n = 3. The claim about im p3 is evident and a computation of the 3 3 linking numbers shows that p3(¯u ) = p3(t¯ ) = 0.
(2,3) for n = 4. The claim about im p4 is evident and a computation of 3 3 the linking numbers shows that p4(E4 \{t¯ , u¯ }) is a base of im p4. One can 0 0 ∼ 0 check that the homomorphism B4 → B4/K4 = B3 maps J4 to J3. Hence ab ab 3 ¯3 ab it induces a homomorphism J4 → J3 which takes u¯ and t of J4 to 3 ¯3 ab 3 ¯3 ab u¯ and t of J3 . Hence rk(ker p4) > rkhu¯ , t i = 2. Since rk J4 = 7 and 3 ¯3 rk(im p4) = 5, we conclude that ker p4 = hu¯ , t i. Remark 2.5. — Note that the braid closures of both u3 and t3 are Bor- romean links. So, maybe, it could be interesting to study how the considered ab base of J3 is related to Milnor’s µ-invariant.
– 1141 – Stepan Yu. Orevkov
2.3. Mixed braid groups and the cabling map
Let n > 1 and ~m = (m1, . . . , mk), m1 + ··· + mk = n, mi ∈ Z, mi > 0. −1 The mixed braid group B ~m (see [19, 20, 10]) is defined as µ (S ~m) where S ~m is the stabilizer of the following vector under the natural action of Sn on Zn: 1,..., 1 , 2,..., 2,...,k,...,k . | {z } | {z } | {z } m1 m2 mk
We emphasize two particular cases: B1,...,1 is the pure braid group and Bn−1,1 is the Artin group corresponding to the Coxeter group of type Bn−1.
We define the cabling map ψ = ψ ~m : Bk × (Bm1 × · · · × Bmk ) → Bn by sending (X; X1,...,Xk) to the braid obtained by replacing each strand of X by a geometric braid representing Xi embedded into a small tubular neighbourhood of this strand. Q Note that ψ ~m is not a homomorphism but its restriction to Pk × Bm Q Q i i is. We have ψ(Pk × i Pmi ) ⊂ Pn and ψ(Pk × i Bmi ) ⊂ B ~m.
ab 3. Jn as an An-module and its automorphisms
Let n > 5. As we mentioned already, by [15, Theorem D], Jn is a charac- 0 0 teristic subgroup of Bn, i.e., Jn is invariant under any automorphism of Bn (in fact a stronger statement is proven in [15]).
0 0 0 Lemma 3.1. — Let ϕ ∈ Aut(Bn) be such that µ ϕ = µ . Let ϕ∗ be the ab automorphism of Jn induced by ϕ|Jn . Then ϕ∗ = ± id. 0 Proof. — The exact sequence 1 → Jn → Bn → An → 1 (see (1.1)) ab 0 0 defines an action of An on Jn by conjugation. The condition µ ϕ = µ implies that ϕ∗ is An-equivariant. Let V be a complex vector space with base e1, . . . , en endowed with the natural action of Sn induced by the action ab 2 on the base. We identify Pn with its image in the symmetric square Sym V by the homomorphism Aij 7→ eiej. Then, by Lemma 2.3, we may identify ab P P Jn with { cijeiej | cij = 0}. These identifications are compatible with the action of An.
For a partition λ = (λ1, . . . , λr), we denote the corresponding irreducible representation of Sn over C (the CSn-module) by Vλ, see, e.g., [7, §4]. For an element v of a CSn-module, let hviCSn be the CSn-submodule generated
– 1142 – Automorphism group of the commutator subgroup of the braid group
⊥ by v. We set e0 = e1 + ··· + en, U = he0iCSn = Ce0, and U = he1 − e2iCSn . 2 Consider the following CSn-submodules of Sym V : 2 P W0 = he1iCSn ,W1 = hwiCSn = Cw where w = i W2 = h(e1 − e2)(e3 + ··· + en)iCSn ,W3 = h(e1 − e2)(e3 − e4)iCSn . We have Sym2 V = Sym2(U ⊕ U ⊥) = Sym2 U ⊕ Sym2 U ⊥ ⊕ (U ⊗ U ⊥) and ⊥ ∼ U = Vn−1,1 (that is Vλ for λ = (n − 1, 1)). It is known (see [17, Lemma 2.1] 2 ∼ ∼ or [7, Exercise 4.19]) that Sym Vn−1,1 = U ⊕ Vn−1,1 ⊕ Vn−2,2 = V ⊕ Vn−2,2. Thus 2 ∼ Sym V = V ⊕ V ⊕ Vn−2,2 . (3.1) ab 2 ∼ Let W = Jn ⊗C. It is clear that Sym V = W0 ⊕W1 ⊕W . Since W0 = V and ∼ ∼ ⊥ W1 = U, we obtain W = U ⊕ Vn−2,2 by cancelling out U ⊕ V in (3.1). Note 2 2 that (e1 −e2)(e3 +···+en) = (e1 −e2)(e0 −(e1 +e2)) = (e1 −e2)e0 −(e1 −e2), 2 2 hence the mapping ei − ej 7→ (ei − ej)e0 − (ei − ej ) induces an isomorphism ⊥ ∼ of CSn-modules U = W2. The identity X (n − 2)(e1 − e2)e3 = (e1 − e2)(e3 + ··· + en) + (e1 − e2)(e3 − ei) (3.2) i>4 shows that W2 + W3 = h(e1 − e2)e3iCSn = W . One easily checks that W2 and W3 are orthogonal to each other with respect to the scalar product on W +W1 for which {eiej}i,j is an orthonormal basis. Therefore W = W2 ⊕W3 is the decomposition of W into irreducible factors. ∼ ∼ We have W2 = Vn−1,1 and W3 = Vn−2,2. Since the corresponding Young diagrams are not symmetric, W2 and W3 are irreducible as CAn-modules (see [7, §5.1]). Since dim W2 6= dim W3 and ϕ∗ is An-equivariant, Schur’s lemma implies that ϕ∗|Wk , k = 2, 3, is multiplication by a constant ck. ab Moreover, since ϕ∗ is an automorphism of Jn (a discrete subgroup), we have ck = ±1. If c3 = −c2 = ±1, then (3.2) contradicts the fact that ab ϕ∗((e1 − e2)e3) ∈ Jn . Let ν ∈ Aut(S6) be defined by (12) 7→(12)(34)(56), (123456) 7→(123)(45). It is well known that ν represents the only nontrivial element of Out(S6). 0 0 0 Lemma 3.2. — Let ϕ ∈ Aut(B6). Then µ ϕ 6= νµ . Proof. — Given a commutative ring k and a kA6-module V correspond- ing to a representation ρ : A6 → GL(V, k), we denote the kA6-module corresponding to the representation ρν by ν∗(V ). It is clear that ν∗ is a covariant functor which preserves direct sums (hence irreducibility), tensor products, symmetric powers etc. 0 0 ab Suppose that µ ϕ = νµ . As in the proof of Lemma 3.1, we endow J6 0 0 with the action of A6. The condition µ ϕ = νµ implies that ϕ induces an – 1143 – Stepan Yu. Orevkov ab ∼ ∗ ab isomorphism of A6-modules J6 = ν (J6 ). Let us show that these modules are not isomorphic. ab ∼ We have J6 ⊗ C = V5,1 ⊕ V4,2 (see the proof of Lemma 3.1). Hence ∗ ab ∼ ∗ ∗ ν (J6 ) ⊗ C = ν (V5,1) ⊕ ν (V4,2). We have dim V5,1 = 5 6= 9 = dim V4,2, ∼ ∗ thus, to complete the proof, it is enough to show that V5,1 =6 ν (V5,1) (note ∼ ∗ that V4,2 = ν (V4,2)). Indeed, ν exchanges the conjugacy classes of the per- mutations a = (123) and b = (123)(456), hence we have χ(a) = 2 6= −1 = χ(b) = χν(a) where χ and χν are the characters of A6 corresponding to V5,1 ∗ and to ν (V5,1) respectively. 4. Centralizers of pure braids Centralizers of braids are computed by González–Meneses and Wiest [10]. For pure braids the answer is much simpler and it can be easily obtained as a specialization of the results of [10]. 4.1. Nielsen–Thurston trichotomy The following definitions and facts we reproduce from [10, Section 2] where they are taken from different sources, mostly from the book [12] which can be also used as a general introduction to the subject. Let D be a disk in C that contains Xn = {1, . . . , n}. The elements of Xn will be called punctures. It is well known that Bn can be identified with the mapping class group D/D0 where D is the group of diffeomorphisms β : D → D such that β|∂D = id∂D and β(Xn) = Xn, and D0 is the connected component of the identity. Sometimes, by abuse of notation, we shall not distinguish between braids and elements of D. For A, B ⊂ D, we write A ∼ B if β0(A) = B for some β0 ∈ D0. An embedded circle in D \ Xn is called an essential curve if it encircles more than one but less than n points of Xn.A multicurve in D \ Xn is a disjoint union of embedded circles. It is called essential if all its components are essential. Let β ∈ D. We say that a multicurve C in D\Xn is stabilized or preserved by β if β(C) ∼ C (the components of C may be permuted by β). The braid represented by β is called reducible if β stabilizes some essential multicurve. A braid β is called periodic if some power of β belongs to Z(Bn). If a braid is neither periodic nor reducible, then it is called pseudo-Anosov; see [12]. – 1144 – Automorphism group of the commutator subgroup of the braid group 4.2. Canonical reduction systems. Tubular and interior braids An essential curve C is called a reduction curve for a braid β if it is stabilized by some power of β and any other curve stabilized by some power of β is isotopic in D\Xn to a curve disjoint from C. An essential multicurve is called a canonical reduction system (CRS) for β if its components represent all isotopy classes of reduction curves for β (each class being represented once). It is known that there exists a canonical reduction system for any braid and that it is unique up to isotopy, see [3], [12, §7], [10, §2]. If a braid is periodic or pseudo-Anosov, the CRS is empty. The following properties of CRS are immediate consequences of their existence and uniqueness. Proposition 4.1. — Let C be the CRS for β ∈ D. Then C is the CRS for β−1. Proposition 4.2. — Let β, γ ⊂ D and let C be the CRS for β. Then γ−1(C) is the CRS for βγ . Proposition 4.3. — Let β, γ ∈ D represent commuting braids. Then: (1) γ preserves the CRS of β. (2) If γ is pure, then it preserves each reduction curve of β. Proof. — (1) follows from Proposition 4.2.(2) follows from (1) We say that a braid is in almost regular form if its CRS is a union of round circles (‘almost’ because the definition of regular form in [10] includes some more conditions which we do not need here). By Proposition 4.2 any braid is conjugate to a braid in almost regular form. Let β be an element of D which represents a reducible braid in almost regular form and let C be a CRS for β. Without loss of generality we may assume that β(C) = C and C is a union of round circles. Let R = R0 ∪ R00 where R0 is the union of the outermost components of C and R00 is the union 0 of small circles around the points of Xn not encircled by curves from R . Let C1,...,Ck be the connected components of R numbered from left to right. Recall that the geometric braid (a union of strings in the cylinder [0, 1] × D) is obtained from β as follows. Let {βt : D → D}t∈[0,1] be an isotopy such that β0 = β, β1 = idD, and βt|∂D = id∂D for any t. Then the i-th string of the geometric braid is the graph of the mapping t 7→ βt(i) and the whole S geometric braid is t {t} × βt(Xn) . Similarly, starting from the circles Ci, S we define the embedded cylinders (tubes) t {t} × βt(Ci) , i = 1, . . . , k. Let mi be the number of punctures encircled by Ci. Following [10, §5.1], we define the interior braid β[i] ∈ Bmi , i = 1, . . . , k, as the element of Bmi corresponding to the union of strings contained in the i-th tube, and we – 1145 – Stepan Yu. Orevkov define the tubular braid βˆ of β as the braid obtained by shrinking each tube to a single string. Let ~m = (m1, . . . , mk) and let ψ ~m be the cabling map (see ˆ Section 2.3). Then we have β = ψ ~m(β; β[1], . . . , β[k]). Recall that C is a CRS for β. Let a be an open connected subset of D such that ∂a ⊂ C ∪∂D. With each such a we associate the braid which is the union of the strings of β starting at a and the strings obtained by shrinking the tubes corresponding to the interior components of ∂a. We denote this braid by β[a]. For example, if a is the exterior component of D \ C, then ˆ β[a] = β. 4.3. Periodic and reducible pure braids The structure of the centralizers of periodic and reducible braids be- comes extremely simple if we restrict our attention to pure braids only. The following fact immediately follows from a result due to Eilenberg [6] and Kerékjártó [13] (see [10, Lemma 3.1]). Proposition 4.4. — A pure braid is periodic if and only if it is a power of ∆2. The following fact can be considered as a specialization of the results of [10]. Proposition 4.5. — Let β be a pure n-braid. (1) If β is periodic, then Z(β; Pn) = Pn. (2) If β is pseudo-Anosov, then Z(β; Pn) is the free abelian group gen- erated by ∆2 and some pseudo-Anosov braid which may or may not coincide with β. (3) If β is reducible non-periodic and in almost regular form, then ψ ~m ˆ maps Z(β; Pk) × Z(β[1]; Pm1 ) × · · · × Z(β[k]; Pmk ) isomorphically onto Z(β; Pn) (see Section 4.2). Proof. — (1). Follows from Proposition 4.4. (2). Follows from [10, Proposition 4.1]. (3). (See also the proof of [10, Proposition 5.17]). By Proposition 4.3 we Q have Z(β; Pn) ⊂ ψ ~m(Pk× Pmi ). The injectivity of the considered mapping −1 and the fact that ψ ~m (Z(β; Pn)) is as stated, are immediate consequences from the following observation: if two geometric braids are isotopic, then the braids obtained from them by removal of some strings are isotopic as well. – 1146 – Automorphism group of the commutator subgroup of the braid group Lemma 4.6. — Let ~m = (m1, . . . , mk), m1 + ··· + mk = n, and p ∈ Z. Then ψ (∆p; ∆p ,..., ∆p ) = ∆p . ~m k m1 mk n Proof. — The result immediately follows from the geometric characteri- zation of ∆ as a braid all whose strings lie on a half-twisted band. Note that the sub-bands of the half-twisted band arising from consecutive strings also consist of half-twisted bands. If X is a periodic pure braid, then X = ∆2d, d ∈ Z, by Proposition 4.4. In this case we set d = deg X, the degree of X. It is clear that lkij(X) = d for any i < j. Lemma 4.7. — Let C be the CRS for a reducible pure braid represented by β ∈ D. Let a and b be two neighboring components of D \ C and let X = β[a] and Y = β[b] be the braids associated with a and b (see the end of Section 4.2). Suppose that each of X and Y is periodic. Then deg X 6= deg Y . 2p 2p Proof. — Suppose that deg X = deg Y = p, i.e., X = ∆k and Y = ∆m for some k, m > 2. Let Ci be the component of C that separates a and b. We may assume that a is exterior to Ci. Let c be the closure of a ∪ b. Then we have 2p 2p 2p β[c] = ψ1,...,1,m,1,...,1(∆k ; 1,..., 1, ∆m , 1,..., 1) = ∆k+m−1 by Lemma 4.6. Hence β[c] preserves any closed curve, in particular a curve which separates some two strings of β[b] and encircles a string of β[c] not belonging to β[b]. Such a curve is not isotopic to any curve disjoint from Ci. This fact contradicts the condition that Ci is a reduction curve. 2 −2 ∼ Lemma 4.8. — Z(σ1σ3 ; Jn) = Pn−2 × Z for n > 4. 2 −2 Proof. — The CRS for σ1σ3 consists of two round circles: one of them encircles the punctures 1 and 2, and the other one encircles the punctures 3 and 4. Then Proposition 4.5(3) implies that ψ = ψ ~m : Pn−2 × (P2 × 2 −2 P2) → Pn, ~m = (2, 2, 1n−4), is injective and im ψ = Z(σ1σ3 ; Pn). One 2 −2 k easily checks that the mapping Pn−2 × P2 → Z(σ1σ3 ; Jn), (X, σ1 ) 7→ k −m ψ(X; σ1 , σ1 ), m = e(ψ(X; 1, 1)) + k, is an isomorphism. Indeed, any ele- 2 −2 k −m ment Y of Z(σ1σ3 ; Pn) is of the form Y = ψ(X; σ1 , σ1 ) and the condition e(Y ) = 0 becomes e(ψ(X; 1, 1)) + k − m = 0. Lemma 4.9. — Let β be a reducible n-braid in almost regular form. Sup- ˆ pose that β ∈ Pk and that the β[i]’s (see Section 4.2) are pairwise non- ˆ conjugate. Then ψ ~m maps Z(β; Pk) × Z(β[1]; Bm1 ) × · · · × Z(β[k]; Bmk ) iso- morphically onto Z(β; Bn) Proof. — See the proof of Proposition 4.5(3). – 1147 – Stepan Yu. Orevkov 5. Proof of Theorem 1.1 for n > 5 −1 5.1. Invariance of the conjugacy class of σ1σ3 0 0 0 Suppose that n > 5. Let ϕ ∈ Aut(Bn) be such that µ ϕ = µ and ϕ∗ = id where ϕ∗ is as in Lemma 3.1. Then we have lki,j(X) = lki,j(ϕ(X)),X ∈ Jn, 1 6 i < j 6 n. (5.1) (n−2)(n−3) −2 Let τ = ψ2,n−2(1; σ1 , ∆ ). We have τ ∈ Jn. 2 −2 Lemma 5.1. — Let X be σ1σ3 or τ. Let α ∈ D represent ϕ(X). Let C be a simple closed curve preserved by α. Suppose that C encircles at least two punctures. Then the punctures 1 and 2 are in the same component of D \ C. Proof. — Suppose that 1 and 2 are separated by C. Without loss of gen- erality we may assume that 1 is outside C and 2 is inside C. Let p be another puncture inside C. Then we have lk1,p(α) = lk1,2(α) which contradicts (5.1) because lk1,2(X) 6= 0 and lk1,p(X) = 0 for any p 6= 2. 2 −2 Lemma 5.2. — Let α ∈ D represent ϕ(σ1σ3 ). Then the CRS for α is invariant under some element of D which exchanges {1, 2} and {3, 4}. Proof. — Follows from Propositions 4.1 and 4.2 because α is conjugate −1 to α and the conjugating element of D exchanges {1, 2} and {3, 4}. Lemma 5.3. — Let α ∈ D represent ϕ(τ). Let C be a component of the CRS for α. Then C cannot separate i and j for all 3 6 i < j 6 n. 2 −2 Proof. — Let β ∈ D represent ϕ(σijσ1 ). Since α and β commute, β preserves C by Proposition 4.3(2). Hence C cannot separate i and j by 2 −2 Lemma 5.1 applied to β (note that β is conjugate to σ1σ3 ; see the beginning of Section 5.2). 2 −2 Lemma 5.4. — Let α ∈ D represent ϕ(σ1σ3 ). Suppose that n > 6. Let C be a component of the CRS for α. Then: (1) C cannot separate 1 and 2. It cannot separate 3 and 4. (2) C cannot separate i and j for 5 6 i < j 6 n. (3) C cannot separate {1, 2, 3, 4} from {5, . . . , n}. (4) C cannot encircle 5, . . . , n. Proof. — (1). Follows from Lemma 5.1 and Lemma 5.2. 2 −2 (2). Let β ∈ D represent ϕ(σijσ1 ). Since α and β commute, β preserves C by Proposition 4.3(2). Hence C cannot separate i and j by Lemma 5.1 applied to β (see the proof of Lemma 5.3). – 1148 – Automorphism group of the commutator subgroup of the braid group (3). Suppose that C separates 1, 2, 3, 4 from 5, 6, . . . , n. Let β ∈ D rep- 2 −2 resent ϕ(σ1σ5 ). Then β is conjugate to α. Let γ ∈ D be a conjugating element. Then γ(C) is a component of the CRS for β and it separates the punctures 1, 2, 5, 6 from all the other punctures. Since α and β commute, β preserves C. This is impossible because the geometric intersection number of C and γ(C) is nonzero. (4). Combine (1), (3), and Lemma 5.2. 2 −2 Lemma 5.5. — Let α ∈ D represent ϕ(σ1σ3 ). Suppose that α is re- ducible non-periodic. Then the CRS for α has exactly two components: one of them encircles 1 and 2, and the other one encircles 3 and 4. Proof. — If n > 6, the result follows from Lemma 5.2 and Lemma 5.4. Suppose that n = 5 and the CRS is not as stated. By combining Lemma 5.2 with Lemma 4.7, we conclude that the CRS consists of a single circle which encircles 1, 2, 3, 4. The interior braid cannot be periodic by (5.1), hence it ∼ 2 is pseudo-Anosov. Therefore, Z(α; P5) = Z by Proposition 4.5(2) whence Z(α; J5) = Z. This contradicts Lemma 4.8. −1 −1 Lemma 5.6. — ϕ(σ1σ3 ) is conjugate in Bn to σ1σ3 . 2 −2 Proof. — Let α ∈ D represent ϕ(σ1σ3 ). If α is pseudo-Anosov, then ∼ 2 Z(α; Pn) = Z by Proposition 4.5(2), hence Z(α; Jn) is abelian which con- tradicts Lemma 4.8. If α is periodic, then it is a power of ∆2 by Proposi- tion 4.4. This contradicts (5.1), hence α is reducible non-periodic and its CRS is as stated in Lemma 5.5. ∼ 2 Suppose that αˆ is pseudo-Anosov. Then Z(ˆα; Pn) = Z by Proposi- ∼ 4 tion 4.5(2) whence Z(α; Pn) = Z by Proposition 4.5(3) and therefore Z(α; Jn) is abelian which contradicts Lemma 4.8. Thus, αˆ is periodic. By Proposition 4.4 this means that αˆ is a power of ∆2. This fact combined 2 −2 2k −2k with (5.1) implies αˆ = 1. It follows that ϕ(σ1σ3 ) is conjugate to σ1 σ3 for some k, and we have k = 1 by (5.1). The uniqueness of roots up to −1 −1 conjugation [9] implies that ϕ(σ1σ3 ) is conjugate to σ1σ3 . Lemma 5.7. — ϕ(τ) is conjugate in Pn to τ. Proof. — Let α ∈ D represent ϕ(τ). By Proposition 4.3, it cannot be −1 pseudo-Anosov because it commutes with ϕ(σ1σ3 ) which is reducible non- periodic by Lemma 5.6. If α were periodic, then it would be a power of ∆2 by Proposition 4.4. This contradicts (5.1), hence α is reducible. Let C be the CRS for α. By Lemmas 5.1 and 5.3, one of the following three cases occurs. Case 1. — C is connected, the punctures 1 and 2 are inside C, all the other punctures are outside C. Then the tubular braid αˆ cannot be pseudo- −1 Anosov because α commutes with ϕ(σ1σ3 ), hence it preserves a circle which – 1149 – Stepan Yu. Orevkov separates 3 and 4 from 5, . . . , n. Hence αˆ is periodic which contradicts (5.1) combined with Proposition 4.4. Thus this case is impossible. Case 2. — C is connected, the punctures 1 and 2 are outside C, all the other punctures are inside C. This case is also impossible and the proof is almost the same as in Case1. To show that αˆ cannot be pseudo-Anosov, we note that α preserves a curve which encircles only 1 and 2. Case 3. — C has two components: c1 and c2 which encircle {1, 2} and {3, . . . , n} respectively. The interior braid α[2] cannot be pseudo-Anosov by the same reasons as in Case1, because α preserves a circle separating 3 and 4 from 5, . . . , n. Hence α[2] is periodic. Using (5.1), we conclude that α is a conjugate of τ. Since the elements of Z(τ; Bn) realize any permutation of {1, 2} and {3, . . . , n}, the conjugating element can be chosen in Pn. Lemma 5.8. — There exists γ ∈ Pn such that −1 −1 γ ϕ(σ1σi ) = (σ1σi ) for i = 3, . . . , n . (5.2) Proof. — Due to Lemma 5.7, without loss of generality we may assume that ϕ(τ) = τ and τ(C) = C where C is the CRS for τ consisting of two round circles c1 and c2 which encircle {1, 2} and {3, . . . , n} respectively. By Lemma 4.9, ψ2,n−2 restricts to an isomorphism ψ : P2 ×B2 ×Bn−2 → Z(τ) := Z(τ; Bn). Let π1 : Z(τ) → P2 and π3 : Z(τ) → Bn−2 be defined as −1 πi = pri ◦ψ . −1 0 −1 Let H = π1 (1) ∩ Bn; note that the elements of π1 (1) correspond to geometric braids whose first two strings are inside the cylinder [0, 1]×c1 and the other strings are inside the cylinder [0, 1]×c2. Then π3|H : H → Bn−2 is −e(X) an isomorphism and its inverse is given by Y 7→ ψ2,n−2(1, σ1 ,Y ), that −1 is σi 7→ σ1 σi+2, i = 1, . . . , n − 3. 0 Let us show that ϕ(H) = H. Indeed, let X ∈ H. Since X ∈ Z(τ; Bn) 0 and ϕ(τ) = τ, we have ϕ(X) ∈ Z(τ; Bn). The fact that π1(X) = 1 follows from (5.1) applied to a power of X belonging to Jn. Hence ϕ(H) ⊂ H. By the same arguments ϕ−1(H) ⊂ H. ∼ Thus ϕ|H is an automorphism of H and we have H = Bn−2. Hence, by Dyer and Grossman’s result [5] cited after the statement of Theorem 1.1, there exists γ ∈ H such that γϕ˜ |H is either idH or Λ|H . The latter case is impossible by (5.1). Thus there exists γ ∈ Bn such that (5.2) holds. It remains to show that γ can be chosen in Pn. By replacing γ with σ1γ if necessary, we may assume that 1 and 2 are fixed by γ. By combining (2.1), (5.1), and (5.2), we conclude that γ({i, j}) = {i, j} for any i, j ∈ {3, . . . , n} and the result follows. – 1150 – Automorphism group of the commutator subgroup of the braid group 5.2. Conjugates of σ1 and simple curves which connect punctures We fix n > 2 and we consider D and the set of punctures Xn = {1, . . . , n} ⊂ D as above. Let I be the set of all smooth simple curves (embedded segments) ◦ ◦ I ⊂ D such that ∂I ⊂ Xn and I ⊂ D \ Xn. Here we denote I = I \ ∂I and ∂I = {a, b} where a and b are the ends of I. Recall that we write I ∼ I1 if I1 = α(I) for some α ∈ D0 (see Section 4.1), i.e., if I and I1 belong to the same connected component of I. Let I ∈ I and let β ∈ D be such that β(I) is the straight line segment β [1, 2]. Then we define the braid σI as σ1 . It is easy to see that σI depends only on the connected component of I that contains I. The CRS for σI is a single closed curve which encloses I and separates it from Xn \ ∂I. By definition, all conjugates of σ1 are obtained in this way. In particular, we have σi = σ[i,i+1] and σij = σI for an embedded segment I which connects i to j passing through the upper half-plane. β Lemma 5.9. — For any β ∈ D, I ∈ I, we have σβ(I) = σI . With this notation, a corollary of Lemma 5.8 can be formulated as follows. 0 Lemma 5.10. — Let n > 5 and let ϕ ∈ Aut(Bn) be as in Section 5.1. (1) Let I,J ∈ I be such that Card(I ∩ J) = Card(∂I ∩ ∂J) = 1 (i.e., I ∩ J is a common endpoint of I and J). Then there exist I1,J1 ∈ I such that I1 ∪ J1 is homeomorphic to I ∪ J and ϕ(σ σ−1) = σ σ−1 , ϕ(σ−1σ ) = σ−1σ . (5.3) I J I1 J1 I J I1 J1 (2) Let I,J ∈ I, I ∩ J = ∅. Then the conclusion is the same as in Part (1). (3) Let I and J be as in Part (1) and let K ∈ I be such that K ∩ (I ∪ J) = ∅. Then there exist I1,J1,K1 ∈ I such that I1 ∪ J1 ∪ K1 is homeomorphic to I ∪ J ∪ K, and (5.3) holds as well as ϕ(σ σ−1) = σ σ−1 , ϕ(σ σ−1) = σ σ−1 . (5.4) K I K1 I1 K J K1 J1 Proof. — (3). Let γ be as in Lemma 5.8 and let β ∈ D be such that −1 β(K) = [1, 2], β(I) = [3, 4], and β(J) = [4, 5]. We set K1 = α (K), I1 = −1 −1 −1 α (I), J1 = α (J) where α = β γϕ(β). Then we have −1 −1 β ϕ(σK σI ) = ϕ((σ1σ3 ) ) by definition of σI and σK −1 γϕ(β) = (σ1σ3 ) by Lemma 5.8 = σ σ−1 by Lemma 5.9 K1 I1 and, similarly, ϕ(σ σ−1) = σ σ−1. Since σ commutes with σ and σ , K J K1 J1 K I J ε −ε −1 −ε −1 ε we have σI σJ = (σK σI ) (σK σJ ) , ε = ±1, thus (5.4) implies (5.3). – 1151 – Stepan Yu. Orevkov (1). Since Card(∂I ∪ ∂J) = 3 and n > 5, we can choose K ∈ I dis- joint from I ∪ J (which is an embedded segment, hence its complement is connected) and the result follows from (3). (2). The same proof as for Part (3) but with β(I) = [1, 2] and β(J) = [3, 4]. Lemma 5.11. — Let I, J, I1,J1 ∈ I be such that I ∩ J = I1 ∩ J1 = ∅. Suppose that σ σ−1 = σ σ−1. Then I ∼ I and J ∼ J . I J I1 J1 1 1 −1 Proof. — It is enough to observe that the CRS for σI σJ is ∂UI ∪ ∂UJ where UI and UJ are ε-neighbourhoods of I and J for 0 < ε 1 (this fact follows, for example, from Lemma 5.5 and Proposition 4.2). Note that when [σI , σJ ] 6= 1, the statement of Lemma 5.11 is wrong. −1 −1 −1 Indeed, in this case by Lemma 5.9 we have σI σJ = σγ(I)σγ(J) for γ = σI σJ whereas σI 6= σγ(I) and σJ 6= σγ(J). Given I,J ∈ I, the geometric intersection number I · J of I and J is ◦ ◦ defined as the minimum of the number of intersection points of I1 and J1 2 over all pairs (I1,J1) ∈ I such that I ∼ I1, J ∼ J1, and I1 is transverse to J1. In this case we say that I1 and J1 realize I · J. p Figure 5.1. Digon removal (p is a puncture) If I,J ∈ I are transverse to each other, we say that a closed embedded disk D is a digon between I and J if D is the closure of a component of D \ (I ∪ J), and ∂D is a union of an arc of I and an arc of J. The common ends of these arcs are called the corners of D. We say that (I0,J 0) is obtained from (I,J) by a digon removal if it is obtained by one of the modifications in Figure 5.1 performed in a neighbourhood of a digon between I and J one of whose corners is not in Xn. The inverse operation is called a digon insertion. The following two lemmas have a lot of analogs in the literature but it is easier to write (and to read) a proof than to search for an appropriate reference. Lemma 5.12. — Let I,J ∈ I be transverse to each other. Then a pair of segments realizing I · J can be obtained from (I,J) by successive digon removals. Proof. — Isotopies of I and of J which transform (I,J) to a pair of seg- ments realizing I · J can be perturbed into a sequence of digon removals and digon insertions. So, it is enough to prove the following “diamond lemma”: if – 1152 – Automorphism group of the commutator subgroup of the braid group (I1,J1) and (I2,J2) are obtained from (I,J) by two different digon removals, then either the pair (I1 ∪ J1,I1) is isotopic to (I2 ∪ J2,I2), or (I1,J1) and (I2,J2) admit digon removals with the same result. We leave it to the reader to check this statement (see Figure 5.2). Figure 5.2. Cases to consider in the diamond lemma 0 0 Lemma 5.13. — Let I1,...,Im ∈ I. Then there exist I1,...,Im ∈ I 0 0 0 such that Ii ∼ Ii for any i = 1, . . . , m, and (Ii,Ij) realizes Ii · Ij for any distinct i, j = 1, . . . , m. Proof. — Induction on the total number of intersection points. If (Ii,Ij) does not realize Ii · Ij, then by Lemma 5.12 there is a digon D between Ii and Ij. We can remove D so that the union of all segments is modified only near the corners of D. Then the total number of intersection points strictly decreases. 5.3. End of the proof Now we are ready to complete the proof of Theorem 1.1 for n > 5. First note that the injectivity of the restriction homomorphism Aut(Bn) → 0 Aut(Bn) is almost evident for any n > 3. Indeed, Let ϕ be an automorphism k of B such that ϕ| 0 = id. By [5], we have ϕ = Λ β˜ with β ∈ B and k = 0 n Bn n 0 k ˜ or 1 (see the introduction). Hence, for any X ∈ Bn, we have Λ β(X) = X, ˜ k n(n−1) −2 i.e., β(X) = Λ (X). In particular, for Xi = σi ∆ , 1 6 i < n, we ˜ have β(Xi) = Xi because Λ(Xi) = Xi. Hence the CRS of each Xi (which is a round circle containing the punctures i and i + 1) is preserved by β; see Proposition 4.3(2). Hence β commutes with all σi for i = 1, . . . , n − 1 k whence β ∈ Z(B ), i.e., β˜ = id. Thus ϕ = Λ . Since Λ| 0 6= id, we conclude n Bn that k = 0, i.e., ϕ = id. 0 Now let us prove that the restriction homomorphism Aut(Bn) → Aut(Bn) 0 is surjective for n > 5. So, let n > 5 and let ϕ be an automorphism of Bn. By [15, Theorem C], we may assume that either µ0ϕ = µ0, or n = 6 and µ0ϕ = νµ0 where ν is as in Section3. However, µ0ϕ 6= νµ0 by Lemma 3.2. So, we assume that µ0ϕ = µ0. Then Lemma 3.1 implies that the automorphism ab ϕ∗ of Jn induced by ϕ is ± id. By composing ϕ with Λ if necessary, we may – 1153 – Stepan Yu. Orevkov assume that ϕ∗ = id (recall that Λ is the automorphism of Bn which takes −1 each σi to σi ). By Lemma 5.8 we may also assume that −1 −1 ϕ(σ1σi ) = σ1σi for all i = 3, . . . , n − 1 (5.5) (otherwise we compose ϕ with γ˜ for the element γ given by Lemma 5.8). Hence −1 −1 −1 −1 ϕ(σiσj ) = σiσj and ϕ(σi σj) = σi σj for all i, j ∈ {3, . . . , n−1}. (5.6) −1 −1 −1 −1 −1 −1 −1 −1 Indeed, σiσj = (σ1σi ) (σ1σj ) and σi σj = (σ1σi )(σ1σj ) . Let I1,J1,K1 ∈ I be as in Lemma 5.10(3) where we set I = [1, 2], J = [2, 3], and K = [4, 5]. By combining (5.4) with (5.5) for i = 4, we obtain σ σ−1 = σ σ−1. Hence I ∼ [1, 2] and K ∼ [4, 5] by Lemma 5.11. Thus, if 1 4 I1 K1 1 1 we set L = J1, then (5.3) reads as −1 −1 −1 −1 ϕ(σ1σ2 ) = σ1σL and ϕ(σ1 σ2) = σ1 σL . (5.7) 2 −2 ∼ By (5.1) for σ2σ4 and by the claim I1 ∪ J1 = I ∪ J of Lemma 5.10(3) we also have ∂L = {2, 3} and L · [1, 2] = 0 . (5.8) By combining (5.5) with (5.7), we obtain −1 −1 −1 −1 ϕ(σiσ2 ) = σiσL and ϕ(σi σ2) = σi σL for all i = 3, . . . , n−1 . (5.9) This fact combined with Lemma 5.10(2) and Lemma 5.11 implies L · [i, i + 1] = 0 for all i = 4, . . . , n − 1 . (5.10) −1 Indeed, for any i = 4, . . . , n − 1, by Lemma 5.10(2) we have ϕ(σ2σi ) = σ σ−1 for some disjoint I ,J ∈ I. On the other hand, ϕ(σ σ−1) = σ σ−1 I1 J1 1 1 2 i L i by (5.9). Hence I1 ∼ L and J1 ∼ [i, i + 1] by Lemma 5.11 whence (5.10) because I1 ∩ J1 = ∅. In the proof of the next lemma for any n > 5, we use Lemma A.1 whose proof is based on Garside theory. However, for n > 6 we also give another proof which uses the material of this section only. Lemma 5.14. — L · [3, 4] = 0. Proof. — By Lemma 5.10(1) applied to I = [2, 3] and J = [3, 4], there exist I ,J such that I ∪J ∼ [2, 4] (thus I ·J = 0) and ϕ(σ σ−1) = σ σ−1. 1 1 1 1 = 1 1 2 3 I1 J1 −1 −1 By combining this fact with (5.9) for i = 3, we obtain σLσ3 = σI1 σJ . γ γ 1 Hence, by Lemma A.1, there exists γ ∈ D such that σL = σI1 and σ3 = σJ1 whence γ(I1) ∼ L and γ(J1) ∼ [2, 3] by Lemma 5.9. Thus L · [3, 4] = I1 · J1 = 0. – 1154 – Automorphism group of the commutator subgroup of the braid group Proof of Lemma 5.14 for n > 6 not using Garside theory. — Let n > 6. We apply the same arguments that we used to obtain (5.7)–(5.9) but we set here I = [3, 4], J = [2, 3], K = [5, 6]. So, let I1,J1,K1 be as in Lemma 5.10(3) for the given choice of I, J, K. By combining (5.4) with (5.6) and (5.9), we obtain σ σ−1 = σ σ−1 and σ σ−1 = σ σ−1. Then Lemma 5.11 yields I = I1 K1 3 5 J1 K1 L 5 1 [3, 4], J1 = L, and K1 = [5, 6]. By Lemma 5.10(3), I1 ∪ J1 is homeomorphic to I ∪ J, hence L · [3, 4] = I · J = 0. Further, Lemma 5.14 combined with (5.8) and (5.10), yields L·[i, i+1] = 0 for any i ∈ {1} ∪ {3, . . . , n − 1}. By Lemma 5.13 this implies that L ∼ L1 where L1 ∈ I is such that [1, 2] ∪ L1 ∪ [3, n] is homeomorphic to a segment. ˜ Hence, up to composing ϕ with β where β ∈ D, β([1, n]) = [1, 2] ∪ L1 ∪ [3, n], we may assume that σL = σ2 in (5.5)–(5.7) and (5.9). This means that ε −ε ε −ε ϕ(σi σj ) = σi σj for any ε = ±1 and any i, j ∈ {1, . . . , n−1}. To complete the proof of Theorem 1.1 for n > 5, it remains to note that the elements ε −ε 0 σi σj , ε = ±1, generate Bn. Indeed, it is shown in [11] (see also [15, §1.8]) 0 −1 −2 −1 −1 that Bn is generated by u = σ2σ1 , v = σ1σ2σ1 = (σ2 σ1)(σ2σ1 ), w = −1 −1 −1 (σ2σ1 )(σ3σ2 ), and ci = σiσ1 , i = 3, . . . , n − 1. Remark 5.15. — Our proof of Theorem 1.1 for n > 5 essentially uses Lemma 5.8 which is based on Dyer–Grossman’s result [5] about Aut(Bn). If n > 6, Lemma 5.8 can be replaced by Lemma A.2 (see below). 6. The case n = 4 0 −1 −1 Recall that B3 is freely generated by u = σ2σ1 and t = σ1 σ2 (see the 0 0 0 Introduction). The group B4 was computed in [11], namely B4 = K4 o B3 where K4 is the kernel of the homomorphism B4 → B3, σ1, σ3 7→ σ1, σ2 7→ −1 −1 σ2. The group K4 is freely generated by c = σ3σ1 and w = σ2c σ2 . The 0 action of B3 on K4 by conjugation is given by ucu−1 = w , uwu−1 = w2c−1w , tct−1 = cw , twt−1 = cw2. (6.1) 0 Besides the elements c, w, u, t of B4, we consider also −1 2 2 3 3 −1 d = ψ2,2(σ1 ; σ1, σ1) = σ1σ3∆ . Lemma 6.1. 2 0 (1) Z(d ; B4) is a semidirect product of infinite cyclic groups hci o hdi where d acts on hci by dcd−1 = c−1. 2 0 (2) hci is a characteristic subgroup of Z(d ; B4). – 1155 – Stepan Yu. Orevkov 2 0 Proof. — Let G = Z(d ; B4). 2 2 (1). We have G = Z(d ; B4) ∩ ker e and, by [10, §5], Z(d ; B4) is the d semidirect product hσ1, σ3i o hdi where d acts on hσ1, σ3i by σ1 = σ3, d σ3 = σ1. (2). Let x be the image of c by an automorphism of G. Then (1) implies that x generates a normal subgroup of G and x is not a power of another −1 element of G. It follows that x ∈ {c, c }. Lemma 6.2. — All the conjugacy classes of B4 which are contained in 0 B4 are presented in Table 6.1. The corresponding centralizers are isomorphic to the groups indicated in this table. 0 0 Proof. — First, note that Bn is normal in Bn, hence for X ∈ Bn, the 0 centralizer Z(X; Bn) depends only on the conjugacy class of X in Bn (though 0 this class may split into several classes in Bn). The centralizers in B4 can be computed by a straightforward applica- tion of [10, Proposition 4.1] and Proposition 4.3. In the computation of 2k+1 −12k−6 Z(∆ σ2 ) (which is, by the way, generated by ∆ and σ2), we use 2k+1 the fact that Z(∆3 ; B3) = h∆i. This fact can be derived either from [10, Proposition 3.5] or from the uniqueness of Garside normal form in B3. In all the cases except, maybe, the following two ones, the computation 0 of Z(X; B4) is evident. k −2 2 Case: X = Y , Y = ψ3,1(σ1 ; ∆3, 1), k 6= 0.— The group Z(X; B4) 2 2 2 is generated by ψ3,1(σ1; 1, 1), σ1, and σ2. Since ψ3,1(σ1; 1, 1) = Y ∆3 and ∆3 ∈ B3, we can choose Y, σ1, σ2 for a generating set, and the result follows because e(Y ) = 0. 2k −6k Case: X = ∆ σ1 , k 6= 0.— The group Z(X; B4) is the isomor- m phic image of B2,1 × Z under the mapping f :(X, m) 7→ ψ2,1(X; σ1 , 1). 0 Hence Z(X; B4) is the isomorphic image of B2,1 under the mapping X 7→ f(X, −e(f(X, 0))). 0 Let ϕ ∈ Aut(B4). ±1 Lemma 6.3. — ϕ(d) is conjugate in B4 to d . 2 0 ∼ 2 0 Proof. — Let x = ϕ(d ). Since Z(x; B4) = Z(d ; B4), we see in Table 6.1 that ϕ(hd2i) ⊂ hd2i. By the same reasons we have ϕ−1(hd2i) ⊂ hd2i, thus ϕ(d2) = d±2 and the result follows from the uniqueness of roots up to con- jugation [9]. – 1156 – Automorphism group of the commutator subgroup of the braid group 0 0 ab CRS XZ(X; B4) Z(X; B4) Z(X; B4) 0 1 B4 B4 Z2 Z −2k 2k 0 3 ψ3,1(σ1 ; ∆3 , 1), k 6= 0 B3 × Z B3 × ZZ Z3 Z2 Z3 Z2 2k 2 d , k 6= 0 Z of Z Z o Z Z × Z2 d2kcl, l 6∈ {0, ±6k} Z3 Z2 d2k+1 Z2 Z 2k −12k 2 ∆ σ1 , k 6= 0 B2,1 × Z B2,1 Z 2k+1 −12k−6 2 ∆ σ2 Z Z Z3 Z2 white/grey region ⇒ the associated braid is periodic/pseudo-Anosov; 2k in Z(d ; B4) we mean f : Z → Aut(Z × Z), f(1)(x, y) = (y, x). 0 Table 6.1. Centralizers of elements of B4 Lemma 6.4. — If ϕ(d) = d, then ϕ(c) = c±1. 2 0 2 0 Proof. — If ϕ(d) = d, then ϕ(Z(d ; B4)) = Z(d ; B4), and we apply Lemma 6.1. 0 Lemma 6.5. — K4 is a characteristic subgroup in B4. Proof. — Lemma 6.3 combined with Lemma 6.4 imply that ϕ(c) is con- jugate to c in B4. Since K4 is the normal closure of c in B4, it follows that ϕ(c) ∈ K4. The same arguments can be applied to any other automorphism 0 of B4, in particular, to ϕσ˜2 whence ϕσ˜2(c) ∈ K4. It remains to recall that ϕσ˜2(c) = ϕ(w) and K4 = hc, wi. Let 1 −1 1 0 S = ,S = , 1 0 1 2 1 1 2 1 1 1 T = S−1S = ,U = S S−1 = . 1 2 1 1 2 1 1 2 Lemma 6.6. — T and U generate a free subgroup of SL(2; Z). Proof. — It is well known that the correspondence σ1 7→ S1, σ2 7→ S2 2 defines an isomorphism B3/h∆ i → PSL(2, Z), see, e.g., [18, §3.5] (this map- ping is also a specialization of the reduced Burau representation). Since – 1157 – Stepan Yu. Orevkov 0 u 7→ U and t 7→ T , the image of B3 = hu, ti is hU, T i. Hence hU, T i is free. Lemma 6.7. — If ϕ|K4 = id, then ϕ = id. 0 0 Proof. — Let ϕ|K4 = id. Since B4 = K4 o B3, we may write ϕ(u) = u1a 0 and ϕ(t) = t1b with u1, t1 ∈ B3 and a, b ∈ K4. For x ∈ K4, we have ϕu˜(x) = −1 −1 −1 ϕ(uxu ) = u1axa u1 =u ˜1a˜(x). Since u˜(x) ∈ K4 and ϕ|K4 = id, we ˜ conclude that u˜(x) = ϕu˜(x) =u ˜1a˜(x). Similarly, t˜(x) = t˜1b(x). Thus, ˜ ˜ ˜ u˜ |K4 =u ˜1b |K4 and t |K4 = t1a˜ |K4 . (6.2) 0 ab Consider the homomorphism π : B4 → Aut(K4 ) = GL(2, Z), x 7→ (˜x)∗; ab here we identify Aut(K4 ) with GL(2, Z) by choosing the images of c and w ab as a base of K4 . It is clear that π(a) = π(b) = 1 and it follows from (6.1) that π(tu−1) = T and π(t) = U. Thus, by Lemma 6.6, the restriction of π 0 to B3 is injective. It follows from (6.2) that π(u1) = π(u) and π(t1) = π(t). Hence u1 = u and t1 = t by the injectivity of π. Then it follows from (6.2) ˜ that a˜ |K4 = b |K4 = id. Since K4 is free, its center is trivial, and we obtain a = b = 1. Thus ϕ = id. Proof of Theorem 1.1 for n = 4.— The injectivity of the restriction 0 homomorphism Aut(B4) → Aut(B4) is already proven in the beginning of 0 Section 5.3, so let us prove the surjectivity. Let ϕ ∈ Aut(B4). By Lemma 6.3, we may assume that ϕ(d) = d±1. Then, by Lemma 6.4, we may assume that ϕ(c) = c±1. Since c∆ = c−1, we may further assume that ϕ(c) = c. By Lemma 6.5, ϕ(c) and ϕ(w) is a free base of K4. Since ϕ(c) = c, it follows p ±1 q that ϕ(w) = c w c , p, q ∈ Z, see [18, §3.5, Problem 3]. We have σ˜1(c) = c −1 and σ˜1(w) = 1231¯2¯1¯ = 1232¯1¯2¯ = 1323¯ 1¯2¯ = c w. (here 1, 1¯, 2,... stand for −1 σ1, σ1 , σ2,... ). Thus, by composing ϕ with a power of c˜ and a power of σ˜1 ±1 ˜ if necessary, we may assume that ϕ(w) = w . For Φ = Λ˜σ1σ˜3∆, we have Φ(c) = c and Φ(w) = 1¯3¯23¯ 1213¯ = 12¯ 3¯221¯ 23¯ = 121¯ 3¯23¯ = 2122¯ 3¯2¯ = w−1 hence, by composing ϕ with Φ if necessary, we may assume that ϕ(c) = c and ϕ(w) = w, thus ϕ|K4 = id and the result follows from Lemma 6.7. Appendix. Garside-theoretic lemmas Here, using Garside theory, we prove two statements one of which (Lem- ma A.1) is used only in the proof of Theorem 1.1 for n = 5, see the proofs of Lemma 5.14, and the other one (Lemma A.2) can be used in the proof of Theorem 1.1 for n > 6 instead of Dyer–Grossman theorem, see Remark 5.15. Let n > 3 and let I and σI ∈ Bn for I ∈ I be as in Section 5.2. – 1158 – Automorphism group of the commutator subgroup of the braid group k l Lemma A.1. — Let k, l ∈ Z \{0} and I,J ∈ I. Suppose that σI σJ is k l u u conjugate to σ1 σ2. Then there exists u ∈ Bn such that σI = σ1 and σJ = σ2, in particular, I · J = 0. Proof. — It follows from Corollary A.4 that there exists u ∈ Bn and u u u p, q, r, s ∈ Z ∩ [1, n] such that σI = σpq and σJ = σrs, This means that σI = u σI1 and σJ = σJ1 where I1,J1 ∈ I satisfy one of the following conditions: (1) I1 ∩ J1 is a common endpoint of I1 and J1; (2) Card(∂I1 ∪ ∂J1) = 4. k l It is enough to exclude condition (2). Indeed, in this case β = σ1 σ2 cannot be conjugate to β = σk σl , because lk (β2) = 0 for i 6∈ {1, 2, 3} and any 1 I1 J1 i,j 2 2 j whereas lkp,q(β1 ) = k and lkr,s(β1 ) = l for pairwise distinct p, q, r, s. Lemma A.2. — Let X and Y be two distinct conjugates of σ1 in Bn, u n > 3. If XYX = YXY , then there exists u ∈ Bn such that X = σ1 and u Y = σ2. Proof. — Follows from Lemma A.5. When speaking of Garside structures on groups, we use the terminology and notation from [21]. Let (G, P, δ) be a symmetric homogeneous square- free Garside structure with set of atoms A (for example, Birman–Ko–Lee’s Garside structure [2] on the braid group, i.e., G = Bn, A = {σij}16i Theorem A.3. — Let k, l ∈ Z \{0}, x, y ∈ A, and let Z = XY where X ∼ xk and Y ∼ yl. Then there exists u ∈ G such that one of the following possibilities holds: u k u l G G (1) X = x1 and Y = y1 with x1 ∈ x ∩ A and y1 ∈ y ∩ A, or (2) `(Zu) = `(Xu) + `(Y u) and Zu ∈ SC(Z). – 1159 – Stepan Yu. Orevkov Proof. — If the statement is true for (k, l), then it is true for (−k, −l), therefore we may assume that l > 0. Then the proof of [21, Corollary 3.5] may be repeated almost word by word in our setting if we define Qm as u u u {Z | u ∈ Um} where Um = {u | `(X ) 6 2m + |k|, `(Y ) = l}. Namely, let m be minimal under the assumption that Qm 6= ∅. If m = 0, then (1) occurs. If m > 0, then, similarly to [21, Lemma 3.3] we show that if u ∈ Um, then `(Zu) = `(Xu) + `(Y u), and similarly to [21, Lemma 3.4] we show that s(Qm) ⊂ Qm. whence Qm ∩ SC(Z) 6= ∅ which implies (2). Corollary A.4. — With the hypothesis of Theorem A.3, assume that Z is conjugate to xkyl. Then there exists u ∈ G such that (1) holds. k l Proof. — Suppose that (2) occurs. Since `(Z) = `s(Z) 6 `(x y ) 6 |k| + u u u |l|, we have `(X )+`(Y ) 6 |k|+|l|. By combining this fact with `(X ) > |k| u u u and `(Y ) > |l|, we obtain `(X ) = |k| and `(Y ) = |l|, and the result follows from [21, Theorem 1a]. Lemma A.5. — Let X ∼ x and Y ∼ y where x, y ∈ A. If XYX = YXY , then there exists u ∈ G such that Xu,Y u ∈ A. Proof. — Without loss of generality we may assume that Y = y ∈ A. −p By [21, Theorem 1a], the left normal form of X is δ ·Ap·...·A1·x·B1·...·Bp i−1 i where Ai,Bi ∈ [1, δ] \{1}, Aiδ Bi = δ for i = 1, . . . , p. By symmetry, the −p right normal form of X is Cp · ... · C1 · x · D1 · ... · Dp · δ again with i−1 i Ci,Di ∈ [1, δ] \{1}, Ciδ Di = δ for i = 1, . . . , p. We have sup yXy 6 2 sup y + sup X = 3 + p. Thus, if p > 0, then sup X + sup y + sup X = 3 + 2p > 3 + p = sup xY x = sup Y xY . Then, by [22, Lemma 2.1b], either Bpy or yCp is a simple element. Without loss of generality we may assume that Bpy ∈ [1, δ]. Since the Garside structure is symmetric and Bpy is simple, there exists −1 v an atom y1 such that Bpy = y1Bp. Thus, for v = Bp , we have y ∈ A and v p−1 the left normal form of X is δ ·Ap−1 ·...·A1 ·x·B1 ·...·Bp−1. Therefore, u u −1 the induction on p yields X = x and Y = z ∈ A for u = (B1 ...Bp) . Remark A.6. — (Compare with [4, 14]). Lemma A.5 admits the fol- lowing generalization which can be proven using the results and methods of [22, 21]. Let X ∼ x and Y ∼ y for x, y ∈ A. Then either X and Y gener- ate a free subgroup of G, or there exists u ∈ G such that Xu,Y u ∈ A. In the latter case, the subgroup generated by X,Y is either free or isomorphic to Artin group of type I2(p), p > 2. In particular, for G = Bn, if X and Y are two conjugates of σ1, then either X and Y generate a free subgroup of Bn, u u or there exists u ∈ Bn such that X = σ1 and Y = σi for some i. Maybe, I will write a proof of this fact in a future paper. – 1160 – Automorphism group of the commutator subgroup of the braid group Bibliography [1] E. Artin, “Theory of braids”, Ann. Math. 48 (1946), p. 101-126. [2] J. Birman, K. H. Ko & S. J. Lee, “A new approach to the word and conjugacy problems in the braid groups”, Adv. Math. 139 (1998), no. 2, p. 322-353. [3] J. Birman, A. Lubotzky & J. McCarthy, “Abelian and solvable subgroups of the mapping class group”, Duke Math. J. 50 (1983), p. 1107-1120. [4] J. Crisp & L. Paris, “The solution to a conjecture of Tits on the subgroup generated by the squares of the generators of an Artin group”, Invent. Math. 145 (2001), no. 1, p. 19-36. [5] J. L. Dyer & E. K. Grossman, “The automorphism groups of the braid groups”, Am. J. Math. 103 (1981), p. 1151-1169. [6] S. Eilenberg, “Sur les transformations périodiques de la surface de sphère”, Funda- menta Math. 22 (1934), p. 28-41. [7] W. Fulton & J. Harris, Representation theory. A first course, Graduate Texts in Mathematics, vol. 129, Springer, 1991, xv+551 pages. [8] V. Gebhardt & J. González-Meneses, “The cyclic sliding operation in Garside groups”, Math. Z. 265 (2010), no. 1, p. 85-114. [9] J. González-Meneses, “The nth root of a braid is unique up conjugacy”, Algebr. Geom. Topol. 3 (2003), p. 1103-1118. [10] J. González-Meneses & B. Wiest, “On the structure of the centralizer of a braid”, Ann. Sci. Éc. Norm. Supér. 37 (2004), no. 5, p. 729-757. [11] E. A. Gorin & V. Lin, “Algebraic equations with continuous coefficients and some problems of the algebraic theory of braids”, Math. USSR-Sbornik 7 (1969), p. 569- 596. [12] N. V. Ivanov, Subgroups of Teichmüller modular groups, Translations of Mathemat- ical Monographs, vol. 115, American Mathematical Society, 1992, xii+127 pages. [13] B. von Kerékjártó, “Über die periodischen Transformationen der Kreisscheibe und der Kugelfläche”, Math. Ann. 80 (1919), p. 36-38. [14] C. J. Leininger & D. Margalit, “Two-generator subgroups of the pure braid group”, Geom. Dedicata 147 (2010), p. 107-113. [15] V. Lin, “Braids and Permutations”, https://arxiv.org/abs/math/0404528, 2004. [16] ——— , “Some problems that I would like to see solved”, Abstract of a talk. Technion, http://www2.math.technion.ac.il/~pincho/Lin/Abstracts.pdf, 2015. [17] K. Magaard, G. Malle & P. H. Tiep, “Irreducibility of tensor squares, symmetric squares and alternating squares”, Pac. J. Math. 202 (2002), no. 2, p. 379-427. [18] W. Magnus, A. Karrass & D. Solitar, Combinatorial group theory: presenta- tions of groups in terms of generators and relations, Pure and Applied Mathematics, vol. 13, Interscience Publishers, 1966, xii+444 pages. [19] S. Manfredini, “Some subgroups of Artin’s braid group”, Topology Appl. 78 (1997), no. 1-2, p. 123-142. [20] S. Yu. Orevkov, “Quasipositivity test via unitary representations of braid groups and its applications to real algebraic curves”, J. Knot Theory Ramifications 10 (2001), no. 7, p. 1005-1023. [21] ——— , “Algorithmic recognition of quasipositive braids of algebraic length two”, J. Algebra 423 (2015), p. 1080-1180. [22] ——— , “Algorithmic recognition of quasipositive 4-braids of algebraic length three”, Groups Complex. Cryptol. 7 (2015), no. 2, p. 157-173. – 1161 –